Let $G$ be a Lie group.

I am wondering if there is a way to say that the map $(g, h)\mapsto dL_g|_h$ defined on $G\times G$ is a smooth map (Here $L_g$ is the left translation map from $G$ to $G$ and by $dL_g|_h$ I mean the differential of $L_g$ at $h$).

The challenge here is to make the set $\bigsqcup_{g, h\in G}\mathcal L(T_hG, T_{gh}G)$ into a smooth manifold in a fruitful way since $dL_g|_h\in \mathcal L(T_gG, T_{gh}G)$.

I was reading the proof of the fact that the Lie algebra of a Lie group is finite dimensional from where the above question is motivated. In the proof there was this fact used that the map $g\mapsto dL_g|_ev:G\to TG$ is a smooth map, where $v$ is a fixed vector in $T_eG$.

Thank you.


1 Answer 1


I think the right way is using pullback bundles.

You need to know in advance that:

$(1)$ pullback bundle is an embedded submanifold of the product of its factors. (Hence it always carries a natural structure of a smooth manifold "inherited from the factors"). For details you can look at this answer.

$(2)$ Given two vector bundles over the same base manifold, you can create their Hom-bundle. The point is that the Hom-bundle can also be given a natural structure of a smooth manifold.

(I do not know an immediate way of showing this, but perhaps it's easier to show this for the tensor product of bundles and use the identification $E^* \otimes F \cong \text{Hom}(E,F)$). Again, as far as I know the literature in this domain is far from optimal).

Assuming for the moment, these two "basic" (and supposedly well-know) facts, we can see the following:

(See here for details). Consider the following two maps $G \times G \to G$:

$m(g,h)=gh \, \,\text{(multiplication)} \, \, , \, \, \pi(g,h)=h$, and their respective pullback bundles: $m^*(TG),\pi^*(TG)$. These are vector bundles over the base space $G \times G$.

Note that fiberwise:


So now if you take their Hom-bundle: $E=\text{Hom}(\pi^*(TG),m^*(TG))$

, then $E$ is a vector bundles over $G \times G$, with the required fiber:


Now your $dL_g|_h$ is just a section of $E$. $E$ is a smooth manifold, so the smoothness of a map $G \times G \to E$ is well defined. You probably want to prove $dL$ is a smooth section of $E$.

$(1)$ You might see a discussion similar in spirit here.

$(2)$ You might also want to consult Lee's book "Introduction to smooth manifolds", page 192 (chapter 8 theorem 8.37), where he proves that every left translation of a vector from $T_eG$ is a smooth vector field.

(He does this without using the language of vector bundles, but it is a very powerful language which is worth knowing anyway...)

  • $\begingroup$ By the way if you are not familiar with vector bundles, pullbacks, Hom-bundles etc, I can recommend some sources. $\endgroup$ Commented Jan 20, 2016 at 22:31
  • $\begingroup$ I know about vector bundles. But I don't know about pullbacks and Hom-bundles. Just have a feeble idea. So it will be great if you could suggest some reading material. $\endgroup$ Commented Jan 21, 2016 at 3:31
  • $\begingroup$ In response to (2). I have been following Lee. And Lee gives a nice proof. I just wanted to see it in a spirit you have discussed. I do not fully understand it yet though. $\endgroup$ Commented Jan 21, 2016 at 3:36
  • $\begingroup$ Also. Can you please also have a look at this. math.stackexchange.com/questions/1334071/… Here I had asked about a basic property of pullbacks. Thanks. $\endgroup$ Commented Jan 21, 2016 at 3:43
  • $\begingroup$ I answered your other question. I suggest you take some time and read it thoroughly. Unfortunately, It's not short. I have also edited my answer here, to stress the point that given the assumptions that Pullback and Hom bundles have natural smooth structures, we can use some iterated uses of these building blocks to build newer (seemingly more complicated) bundles. (Which now get their own smooth structure) $\endgroup$ Commented Jan 21, 2016 at 11:57

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